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2 years ago

6

The music is in ___meter? a.free timeb.duplec.tripled.quadruple

2 answers:

KonstantinChe [14]2 years ago

6 0

Answer:

letter b. duple

Explanation:

anong fav subject mo hehe curious lang ako ehh pa brainly naman plsss

Rashid [163]2 years ago

4 0

D.quadruple is the answer

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What current is needed in the solenoid's wires?

marta [7]

Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.

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2 years ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

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2 years ago

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worty [1.4K]

Answer:

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Explanation:

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Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

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Hence the magnification is 2.74

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